The plane in space, various equations, the distance from a point to a plane.

There are such forms of writing the equation of a plane:

1) $Ax+By+Cz+D=0-$ the general equation of the plane $P,$ where $\bar{N}=(A,B,C)-$ the normal vector of the plane $P.$

2) $A(x-x_0)+B(y-y_0)+C(z-z_0)=0-$ the equation of the plane $P,$ which passes through the point $M(x_0,y_0,z_0)$ perpendicular to the vector $\bar{N}=(A,B,C).$ The vector $\bar{N}$ is called the normal vector of the plane.

3) $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1-$ the equation of the plane in intercepts on the axes, where $a,$ $b,$ and $c-$ the lengths of the segments that the plane cuts off on the coordinate axes.

4) $$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& x_2-x_1& x_3-x_1\end{array}\right|=0$$ - the equation of the plane, which passes through three points $A(x_1,y_1,z_1),B(x_2,y_2,z_2),$ and $C(x_3,y_3,z_3).$

5) $x\cos \alpha +y\cos \beta +z\cos \gamma -p=0-$ the normal equation of the plane, where $\cos \alpha ,\cos \beta ,$ and $\cos \gamma -$ the directional cosines of the normal vector $\bar{N},$ directed from the origin towards the plane, and $p>0-$ the distance from the origin to the plane.

The general equation of the plane is converted to the normal form by multiplying by the normalizing factor $\mu =-\frac{sgnD}{\sqrt{A^2+B^2+C^2}}.$

The distance from the point $M(x_0,y_0,z_0)$ to the plane $P:Ax+By+Cz+D=0$ is calculated using the formula. $$d=\left|\frac{A{x}_0+B{y}_0+C{z}_0+D}{\sqrt{A^2+B^2+C^2}}\right|.$$

Examples:

1.

a) Given the plane $P:-2x+y-z+1=0$ and the point $M(1,1,1).$ Write the equation of the plane $P^{\prime },$ passing through the point $M$ parallel to the plane $P$ and calculate the distance $\rho (P,P^{\prime }).$

Solution.

Since the planes $P$ and $P^{\prime }$ are parallel, the normal vector for the plane $P$ will also be the normal vector for the plane $P^{\prime }.$ From the equation of the plane, we get $\bar{N}=(-2,1,-1).$

Next, we write the equation of the plane using the formula (2): $A(x-x_0)+B(y-y_0)+C(z-z_0)=0-$ the equation of the plane that passes through the point $M(x_0,y_0,z_0)$ perpendicular to the vector $\bar{N}=(A,B,C).$

$-2(x-1)+(y-1)-(z-1)=0\Rightarrow -2x+y-z+2=0.$

Answer: $-2x+y-z+2=0.$

2.

a) Write the equation of the plane $P^{\prime },$ passing through the given points $M_1(1,2,0)$ and $M_2(2,1,1)$ perpendicular to the given plane $P:-x+y-1=0.$

Solution.

From the equation of the plane $P,$ we find its normal vector $\bar{N}=(-1,1,0).$ A plane perpendicular to plane $P$ is parallel to its normal vector. It follows that we can choose a point $M_3(x,y,z)\in P^{\prime }$ such that $\overline{M_1{M}_3}||\bar{N}.$

$\overline{M_1{M}_3}=(x-1,y-2,z).$

The condition for the collinearity of vectors $\overline{M_1{M}_3}$ and $\bar{N}:$ $\frac{x_{M_1{M}_3}}{x_N}=\frac{y_{M_1{M}_3}}{y_N}=\frac{z_{M_1{M}_3}}{z_N}.$

Since $z_N=0,$ i.e., the vector $N$ lies in the $XoY$ plane, $z_{M_1{M}_3}=0.$

$\frac{x-1}{-1}=\frac{y-2}{1}.$ Let $x=2,$ then $y=1.$

We found the point $M_3=(2,1,0).$

Since point $M_1\in P^{\prime },$ then $M_3\in P^{\prime }$ as well. Write the equation of the plane passing through three points $M_1(1,2,0),$ $M_2(2,1,1),$ and $M_3(2,1,0).$

$\left|\begin{array}{ccc}x-1& y-2& z\\ 1& -1& 1\\ 1& -1& 0\end{array}\right|=0\Rightarrow$

$(x-1)(-1)0+(-1)z+(y-2)-(-1)z-(-1)(x-1)-(y-2)0=0\Rightarrow$ $\Rightarrow -z+y-2+z+x-1=0\Rightarrow x+y-3=0.$

Answer: $x+y-3=0.$

3.

a) Write the equation of the plane $P,$ passing through the point $M(1,1,1)$ parallel to vectors $a_1(0,1,2)$ and $a_2(-1,0,1).$

Solution.

Since the vector $[a_1,a_2]$ is perpendicular to the plane of vectors $a_1$ and $a_2$ (see vector product), it will also be perpendicular to the desired plane. Thus, the vector $[a_1,a_2]$ is normal to the plane $P.$ Let's find this vector:

$[a_1,a_2]=\left|\begin{array}{ccc}i& j& k\\ 0& 1& 2\\ -1& 0& 1\end{array}\right|=i(1-0)-j(0+2)+k(0+1)=i-2j+k.$

Hence, $\bar{N}=[a_1,a_2]=(1,-2,1).$

Now we can find the equation of the plane $P,$ using the formula (2), as the plane passing through the point $M(1,1,1)$ perpendicular to the vector $\bar{N}=(1,-2,1):$

$1(x-1)-2(y-1)+1(z-1)=0\Rightarrow$

$x-2y+z=0.$

Answer: $x-2y+z=0.$

4.

a) Write the equation of the plane $P,$ passing through points $M_1(1,2,0)$ and $M_2(2,1,1)$ parallel to the vector $a=(3,0,1).$

Solution.

Since the vector $a$ is parallel to the plane $P,$ for any vector $\overline{M_1{M}_3},$ parallel to vector $a,$ the point $M_3$ lies in $P.$

Let $M_3=(x,y,z).$ Then $\overline{M_1{M}_3}=(x-1,y-2,z).$ As $\overline{M_1{M}_3}\parallel a,$ we have $\frac{x_{M_1{M}_3}}{x_a}=\frac{y_{M_1{M}_3}}{y_a}=\frac{z_{M_1{M}_3}}{z_a}.$ Since $y_a=0,$ vector $a$ lies in the $XoZ$ plane, and any vector parallel to it will also lie in this plane. Hence, $y_{M_1{M}_3}=y-2=0\Rightarrow y=2.$

From the condition of parallelism of vectors, we have $\frac{x-1}{3}=\frac{z}{1}.$ Let's take $x=4,$ then $z=1.$

We have found the point $M_3=(4,2,1).$

Now, write the equation of the plane passing through three points $M_1(1,2,0),M_2(2,1,1),$ and $M_3(4,2,1).$

$\left|\begin{array}{ccc}x-1& y-2& z\\ 1& -1& 1\\ 3& 0& 1\end{array}\right|=0\Rightarrow$

$(x-1)(-1)1+1\cdot z\cdot 0+(y-2)3-3(-1)z-0\cdot 1\cdot (x-1)-1(y-2)1=0\Rightarrow$

$-x+1+3y-6+3z-y+2=0\Rightarrow -x+2y+3z-3=0.$

Answer: $-x+2y+3z-3=0.$

5.

a) Write the equation of the plane passing through three given points $M_1(1,2,0),$ $M_2(2,1,1)$ and $M_3(3,0,1).$

Solution.

Let's use formula (4):

$\left|\begin{array}{ccc}x-1& y-2& z\\ 2-1& 1-2& 1\\ 3-1& 0-2& 1\end{array}\right|=0\Rightarrow$

$\left|\begin{array}{ccc}x-1& y-2& z\\ 1& -1& 1\\ 2& -2& 1\end{array}\right|=0\Rightarrow$

$(x-1)(-1)\cdot 1+z(-2)+2(y-2)\cdot 1-2(-1)z-(-2)(x-1)-1(y-2)\cdot 1=0\Rightarrow$

$\Rightarrow -x+1-2z+2y-4+2z+2x-2-y+2=0\Rightarrow x+y-3=0.$

Answer: $x+y-3=0.$

Tags: plane in space, distance, geometry